Q: A long solenoid of 50\,\text{cm} length having 100 turns carries a current of 2.5\,\text{A}. The magnetic field at the centre of solenoid is (Take, \mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1})

6.28\times10^{-4}\,\text{T}. B=\mu_0 n I=\mu_0\left(\dfrac{N}{l}\right)I=4\pi\times10^{-7}\times\dfrac{100}{0.5}\times2.5=6.28\times10^{-4}\,\text{T}.

Q: Two wires are held perpendicular to the plane of paper and are 5\,\text{m} apart. They carry currents of 2.5\,\text{A} and 5\,\text{A} in the same direction. Then, the magnetic field strength (B) at a point midway between the wires will be

\dfrac{\mu_0}{2\pi}\,\text{T}. At the midpoint (2.5 m from each wire), the two fields point opposite to each other: B_1=\dfrac{\mu_0(2.5)}{2\pi(2.5)}=\dfrac{\mu_0}{2\pi}, B_2=\dfrac{\mu_0(5)}{2\pi(2.5)}=\dfrac{\mu_0}{\pi}. Net B=B_2-B_1=\dfrac{\mu_0}{2\pi}\,\text{T}.

Question 1

A long solenoid of 50\,	ext{cm} length having 100 turns carries a current of 2.5\,	ext{A}. The magnetic field at the centre of solenoid is (Take, \mu_0=4\pi	imes10^{-7}\,	ext{T m A}^{-1})

Accepted Answer

6.28	imes10^{-4}\,	ext{T}. B=\mu_0 n I=\mu_0\left(\dfrac{N}{l}ight)I=4\pi	imes10^{-7}	imes\dfrac{100}{0.5}	imes2.5=6.28	imes10^{-4}\,	ext{T}.

Question 2

Two toroids 1 and 2 have total number of turns 200 and 100 respectively with average radii 40 cm and 20 cm respectively. If they carry same current i, the ratio of the magnetic fields along the two loops is

Accepted Answer

1:1. For a toroid, B=\dfrac{\mu_0 Ni}{2\pi r}. So \dfrac{B_1}{B_2}=\dfrac{N_1}{N_2}	imes\dfrac{r_2}{r_1}=\dfrac{200}{100}	imes\dfrac{20}{40}=1, giving B_1:B_2=1:1.

Question 3

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' at radial distances \dfrac{a}{2} and 2a respectively, from the axis of the wire is

Accepted Answer

1. Inside: B=\dfrac{\mu_0 I r}{2\pi a^2}, at r=a/2: B=\dfrac{\mu_0 I}{4\pi a}. Outside: B'=\dfrac{\mu_0 I}{2\pi r}, at r=2a: B'=\dfrac{\mu_0 I}{4\pi a}. So B:B'=1:1.

Question 4

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

Accepted Answer

n^2B. Since the wire length is fixed, the radius becomes R/n for n turns. B_{centre}=\dfrac{n\mu_0 i}{2r}; substituting r=R/n gives B_{new}=n^2 B.

Question 5

An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude

Accepted Answer

\dfrac{\mu_0 n e}{2r}. Equivalent current I=\dfrac{q}{T}=ne (since T=1/n). Magnetic field at centre B=\dfrac{\mu_0 I}{2r}=\dfrac{\mu_0 n e}{2r}.

Question 6

Two identical long conducting wires AOB and COD are placed at right angles to each other, such that one is above the other and O is their common point. The wires carry I_1 and I_2 currents, respectively. Point P is lying at distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be

Accepted Answer

\dfrac{\mu_0}{2\pi d}(I_1^2+I_2^2)^{1/2}. Fields B_1=\dfrac{\mu_0 I_1}{2\pi d} and B_2=\dfrac{\mu_0 I_2}{2\pi d} are mutually perpendicular, so B_{net}=\sqrt{B_1^2+B_2^2}=\dfrac{\mu_0}{2\pi d}(I_1^2+I_2^2)^{1/2}.

Question 7

When a proton is released from rest in a room, it starts with an initial acceleration a_0 towards West. When it is projected towards North with a speed v_0, it moves with an initial acceleration 3a_0 towards West. The electric and magnetic fields in the room are

Accepted Answer

\dfrac{ma_0}{e} West, \dfrac{2ma_0}{ev_0} down. From rest, only the electric force acts: a_0=\dfrac{eE}{m}\Rightarrow E=\dfrac{ma_0}{e} West. Moving North with v_0, the magnetic force adds: ev_0B+eE=3ma_0\Rightarrow B=\dfrac{2ma_0}{ev_0}, directed vertically downward.

Question 8

Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be

Accepted Answer

\dfrac{\sqrt5\,\mu_0 I}{2R}. Fields at centre are B_1=\dfrac{\mu_0 I}{2R} and B_2=\dfrac{\mu_0(2I)}{2R}=2B_1, mutually perpendicular. Resultant =\sqrt{B_1^2+B_2^2}=\sqrt5\,B_1=\dfrac{\sqrt5\,\mu_0 I}{2R}.

Question 9

Two wires are held perpendicular to the plane of paper and are 5\,	ext{m} apart. They carry currents of 2.5\,	ext{A} and 5\,	ext{A} in the same direction. Then, the magnetic field strength (B) at a point midway between the wires will be

Accepted Answer

\dfrac{\mu_0}{2\pi}\,	ext{T}. At the midpoint (2.5 m from each wire), the two fields point opposite to each other: B_1=\dfrac{\mu_0(2.5)}{2\pi(2.5)}=\dfrac{\mu_0}{2\pi}, B_2=\dfrac{\mu_0(5)}{2\pi(2.5)}=\dfrac{\mu_0}{\pi}. Net B=B_2-B_1=\dfrac{\mu_0}{2\pi}\,	ext{T}.

Question 10

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same?

Accepted Answer

4. B=\dfrac{\mu_0 i}{2r}; equal B needs \dfrac{i_1}{r_1}=\dfrac{i_2}{r_2}, and with r_1=2r_2, i_1=2i_2. Since resistance \propto length \propto r, \dfrac{V_1}{V_2}=\dfrac{i_1 r_1}{i_2 r_2}=\dfrac{2i_2(2r_2)}{i_2 r_2}=4.

Question 11

A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is

Accepted Answer

B. For a solenoid, B=\mu_0 n i. With n_2=n/2 and i_2=2i, B_2=\mu_0\left(\dfrac{n}{2}\right)(2i)=\mu_0 n i=B, so the field stays unchanged.

Question 12

The magnetic field of a given length of wire carrying a current for a single turn circular coil at centre is B, then its value for two turns for the same wire when same current passing through it is

Accepted Answer

4B. For N turns made from the same wire of length L, the radius r=\dfrac{L}{2\pi N}, so B=\dfrac{\mu_0 N i}{2r}=\dfrac{\mu_0 N^2 i}{L}\propto N^2. Doubling the turns therefore gives B_2=4B.

Question 13

Magnetic field due to 0.1\,	ext{A} current flowing through a circular coil of radius 0.1\,	ext{m} and 1000 turns at the centre of the coil is

Accepted Answer

6.28	imes10^{-4}\,	ext{T}. B=\dfrac{\mu_0 N i}{2r}=\dfrac{4\pi	imes10^{-7}	imes1000	imes0.1}{2	imes0.1}=2\pi	imes10^{-4}=6.28	imes10^{-4}\,	ext{T}.

Question 14

If a long hollow copper pipe carries a current, then magnetic field is produced

Accepted Answer

outside the pipe only. By Ampere's circuital law \oint \vec{B}\cdot d\vec{l}=\mu_0 i_{	ext{enclosed}}; inside the hollow pipe i_{	ext{enclosed}}=0 so B=0, while outside the pipe the current behaves as if concentrated on the axis, giving B=\dfrac{\mu_0 i}{2\pi r}
eq 0.

Question 15

A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be

Accepted Answer

1:4. Since B\propto \dfrac{N^2}{L} for a fixed wire length L, \dfrac{B_1}{B_2}=\dfrac{N_1^2}{N_2^2}=\dfrac{1^2}{2^2}=\dfrac{1}{4}.

Question 16

The magnetic field d\vec{B} due to a small element d\vec{l} at a distance \vec{r} and carrying current i is

Accepted Answer

d\vec{B}=\dfrac{\mu_0}{4\pi}i\left(\dfrac{d\vec{l}	imes\vec{r}}{r^3}ight). By the Biot-Savart law, dB\propto i,\, dl,\, \sin	heta and dB\propto 1/r^2; combining these with the cross product d\vec{l}	imes\vec{r} (which already carries the \sin	heta factor) gives d\vec{B}=\dfrac{\mu_0}{4\pi}\,i\,\dfrac{d\vec{l}	imes\vec{r}}{r^3}.

Question 17

At what distance from a long straight wire carrying a current of 12\,	ext{A} will the magnetic field be equal to 3	imes10^{-5}\,	ext{Wb/m}^2?

Accepted Answer

8	imes10^{-2}\,	ext{m}. B=\dfrac{\mu_0}{2\pi}\dfrac{i}{r}\Rightarrow r=\dfrac{\mu_0 i}{2\pi B}=\dfrac{4\pi	imes10^{-7}	imes12}{2\pi	imes3	imes10^{-5}}=8	imes10^{-2}\,	ext{m}.

Question 18

A straight wire of diameter 0.5\,	ext{mm} carrying a current of 1\,	ext{A} is replaced by another wire of 1\,	ext{mm} diameter carrying the same current. The strength of magnetic field far away is

Accepted Answer

same as the earlier value. The field far from a straight wire, B=\dfrac{\mu_0}{4\pi}\dfrac{2i}{r}, depends only on the current i and the distance r from the wire, not on the wire's diameter, so the field is unchanged.

Question 19

The magnetic field at a distance r from a long wire carrying current i is 0.4\,	ext{T}. The magnetic field at a distance 2r is

Accepted Answer

0.2\,	ext{T}. Since B\propto \dfrac{1}{r}, doubling the distance halves the field: B'=\dfrac{0.4}{2}=0.2\,	ext{T}.

Question 20

The magnetic induction at a point P which is at a distance of 4\,	ext{cm} from a long current carrying wire is 10^{-3}\,	ext{T}. The field of induction at a distance 12\,	ext{cm} from the current will be

Accepted Answer

3.33	imes10^{-4}\,	ext{T}. Since B\propto \dfrac{1}{r}, tripling the distance from 4\,	ext{cm} to 12\,	ext{cm} reduces the field to one-third: B'=\dfrac{10^{-3}}{3}=3.33	imes10^{-4}\,	ext{T}.

Biot Savart's Law and Ampere's Circuital Law — NEET Physics PYQ MCQs with Solutions

Questions with solutions